%I #21 Apr 15 2021 04:03:48
%S 1,3,11,31,71,145,289,601,1321,2979,6683,14743,32111,69697,151777,
%T 332113,728689,1598883,3503627,7668079,16774775,36704017,80343361,
%U 175916521,385196761,843365379,1846290395,4041672871,8847607391,19368919297,42403014721,92830645537
%N a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) + a(n-4).
%C a(n)/a(n-1) tends to 2.189207115... = 1 + 2^(1/4) = 1 + A010767.
%D E. J. Barbeau, Polynomials, Springer-Verlag NY Inc, 1989, p. 136.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,1).
%F We use a 4 X 4 matrix corresponding to the characteristic polynomial (x - 1)^4 - 2 = 0 = x^4 - 4x^3 + 6x^2 - 4x - 1 = 0, being [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / 1 4 -6 4]. Let the matrix = M. Perform M^n * [1, 1, 1, 1]. a(n) = the third term from the left, (the other 3 terms being offset members of the series).
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)+a(n-4). G.f.: -x*(x^3+5*x^2-x+1)/ (x^4+4*x^3-6*x^2+4*x-1). [_Colin Barker_, Oct 21 2012]
%e a(4) = 31, since M^4 * [1,1,1,1] = [3, 11, 31, 71].
%t LinearRecurrence[{4,-6,4,1},{1,3,11,31},40] (* _Harvey P. Dale_, Jul 22 2013 *)
%Y Cf. A010767, A052101.
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Mar 28 2004
%E Corrected by _T. D. Noe_, Nov 08 2006
%E New name using recurrence from _Colin Barker_, _Joerg Arndt_, Apr 15 2021